2010 H2 Mathematics Paper 2 Question 2

Sigma Notation

Answers

(i)

Out of syllabus
(iib)
As n→∞,{ n \to \infty, } 12(n+1),12(n+2)→0{\frac{1}{2 \left( n + 1 \right)},\frac{1}{2 \left( n + 2 \right)} \to 0} so 34−12(n+1)−12(n+2)→34{\frac{3}{4}-\frac{1}{2(n+1)}-\frac{1}{2(n+2)} \to \frac{3}{4}}
Hence the series is convergent
Sum to infinity =34{= \frac{3}{4}}

Full solutions

(i)

Out of syllabus
(iia)
∑r=1n1r(r+2)=∑r=1n(12r−12(r+2))=12−16+14−18+16−110+⋯+12(n−2)−12n+12(n−1)−12(n+1)+12n−12(n+2)=34−12(n+1)−12(n+2)  ■\begin{align*} & \sum_{r=1}^n \frac{1}{r(r+2)} \\ &= \sum_{r=1}^n \left(\frac{1}{2 r} - \frac{1}{2 \left( r + 2 \right)}\right) \\ & = \def\arraystretch{1.5} \begin{array}{lclc} & \frac{1}{2} &-& \cancel{\frac{1}{6}} \\ + & \frac{1}{4} &-& \cancel{\frac{1}{8}} \\ + & \cancel{\frac{1}{6}} &-& \cancel{\frac{1}{10}} \\ + & & \cdots & \\ + & \cancel{\frac{1}{2 \left( n - 2 \right)}} &-& \cancel{\frac{1}{2 n}} \\ + & \cancel{\frac{1}{2 \left( n - 1 \right)}} &-& \frac{1}{2 \left( n + 1 \right)} \\ + & \cancel{\frac{1}{2 n}} &-& \frac{1}{2 \left( n + 2 \right)} \end{array} \\ \\ &= \frac{3}{4} - \frac{1}{2 \left( n + 1 \right)} - \frac{1}{2 \left( n + 2 \right)} \; \blacksquare \end{align*}
(iib)
As n→∞,{ n \to \infty, } 12(n+1),12(n+2)→0{\displaystyle \frac{1}{2 \left( n + 1 \right)},\frac{1}{2 \left( n + 2 \right)} \to 0} so
∑r=1n1r(r+2)=34−12(n+1)−12(n+2)→34\begin{align*} & \sum_{r=1}^n \frac{1}{r(r+2)} \\ &= \frac{3}{4}-\frac{1}{2(n+1)}-\frac{1}{2(n+2)} \\ &\to \frac{3}{4} \end{align*}
Hence ∑r=1∞1r(r+2){\displaystyle \sum_{r=1}^\infty \frac{1}{r(r+2)}} is convergent ■{\blacksquare}
Sum to infinity=∑r=1∞1r(r+2)=34  ■\begin{align*} &\textrm{Sum to infinity} \\ &= \sum_{r=1}^\infty \frac{1}{r(r+2)} \\ &= \frac{3}{4} \; \blacksquare \end{align*}